Question

A body is projected such that the horizontal range is twice the maximum height of projection. Find the angle of projection; and the time of flight if the velocity of projection is 5m/s

Answer 1

vertical ... 2[5 sin(θ)] / g = t
... h = 5/2 sin(θ) * t/2

horizontal ... d = 5 cos(θ) t

2h = d
5/2 sin(θ) t = 5 cos(θ) t
dividing by cos(θ) t
and multiplying by 2/5
tan(θ) = 2

substitute back to find t

Answer 2

I don't really get you scott

Answer 3

Am I to substitute it this way 2[5(sin2)]/9.8.? Or will I make use of tan instead, since tan=2

Answer 4

Do tan-1 to find theta (how do you guys make a theta on this thing). Sub that back into the very first eq to find t